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So I have this homework. Note that $J_\epsilon$ is the standard mollifier.

If $\Omega\in\mathbb{R}^n$ open and $u\in C(\Omega)$, show that $J_\epsilon * u\rightarrow u$ uniformly on every compact subset of $\Omega$ as $\epsilon\rightarrow0$.

At some point we were given a hint that $J_\epsilon * u(x)-u(x)=\int J_\epsilon(y) t(u(x-y)-u(x)) dy$. After that since $u$ is uniformly continuous on compact sets, we may choose one such set and then bound $|J_\epsilon * u(x)-u(x)|$ by a constant. My problem is that I cannot see how we get the first equality, doesn't it require a constant $J_\epsilon * u(x)-u(x)=\int J_\epsilon(y) t(u(x-y)-C u(x)) dy$?

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Prove that $\int J_\varepsilon =1$ for every $\varepsilon$. –  Giuseppe Negro Jan 14 '13 at 14:39
    
Oh, yes, this is by definition and I am stupid. It is very clear now. Thanks –  rom Jan 14 '13 at 14:58

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I think you misplaced a '$t$' in your equality and that you should integrate with respect to $y$: \begin{align} J_\varepsilon\ast u(x) - u(x) = & \int_{\mathbb{R}^n}\!\!\!\!J_\varepsilon(y)u(x-y){\rm d}y ~-~ u(x) \\ = & \int_{\mathbb{R}^n}\!\!\!\!J_\varepsilon(y)u(x-y){\rm d}y ~-~ u(x)\int_{\mathbb{R}^n}\!\!\!\!J_\varepsilon(y){\rm d}y \\ = & \int_{\mathbb{R}^n}\!\!\!\!J_\varepsilon(y)\Big(u(x-y)-u(x)\Big){\rm d}y \end{align} Remember that by definition a mollifier $\{J_\varepsilon\}_{\varepsilon>0}$ satisfies

  1. $J_\varepsilon\in C^\infty(\mathbb{R}^n)$;
  2. $J_\varepsilon(x) = 0~~$ $\forall x\in\mathbb{R}^n\setminus {\rm B}(0,\varepsilon)$;
  3. $J_\varepsilon$ is a probability measure on $\mathbb{R}^n$, i.e. $J_\varepsilon\geq 0$ and $\displaystyle{\int_{\mathbb{R}^n}\!\!\!\!J_\varepsilon = 1}$. The second equality above follows from this last property.
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Yes, the integration should be done with respect to $y$. –  rom Jan 14 '13 at 16:03

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